Suppose that \(A \subseteq \mathbf {H}\)  has zero Lebesgue measure. We want to show that \(\mathbf {E}[\mu ^0](A) = 0\) . It suffices to show that \(\mathbf {E}[ \mu ^0](A \cap (\mathbf {H}+ ir)) =0\)  for each \(r > 0\) . We may therefore assume that \(A \subseteq \mathbf {H}+ir\) . We will prove the result in the case that \(r=2\) . The result for other values of \(r > 0\)  follows from the same argument. By [1]} we have that \(2 = \mathrm {hcap}(\eta ([0,1])) \ge \sup \lbrace \mathrm {Im}(z) : z \in \eta ([0,1])\rbrace ^2/2\) . That is, \(\sup \lbrace \mathrm {Im}(z) : z \in \eta ([0,1])\rbrace  \le 2\) . Therefore \(f_t(z)\)  is defined for all \(z \in A\)  and \(t \in [0,1]\) . Let \(U\)  be uniform in \([0,1]\)  independently of everything else. By (REF ) we have that
\( \mathbf {E}[ \mu ^0](A) = \mathbf {E}\left[ \int |\psi _U^{\prime }(z)|^{d} \mathbf {1}_{z \in f_U(A)} d\mu ^U(z) \right] = \int \mathbf {E}[ |\psi _U^{\prime }(z)|^{d} \mathbf {1}_{z \in f_U(A)}] d\mathbf {E}[\mu ^0(z)]\) 
where in second equality we used that \(\mu ^U \stackrel{d}{=} \mu ^0\)  and that \(\mu ^U\)  is independent of \(f_U\) . To show that the right hand side is equal to zero, it suffices to show that \(\mathbf {P}[ \psi _U(z) \in A] = 0\)  for each \(z \in \mathbf {H}\) . Let \((\widetilde{f}_t)\)  solve the centered reverse Loewner equation with driving function \(\widetilde{W}_t = \sqrt{\kappa } \widetilde{B}_t\)  where \(\widetilde{B}\)  is a standard Brownian motion. That is,
\( d\widetilde{f}_t(z) = -\frac{2}{\widetilde{f}_t(z)} dt - d\widetilde{W}_t,\quad \widetilde{f}_0(z) = z.\) 
Then we have that \(\psi _t(z) \stackrel{d}{=} \widetilde{f}_t(z)\)  for all \(t\) . Consequently, it suffices to show that \(\mathbf {P}[ \widetilde{f}_U(z) \in A] = 0\)  for each \(z \in \mathbf {H}\) . Let \(\widetilde{Z}_t = \widetilde{f}_t(z)\) , \(\widetilde{\Theta }_t = \arg \widetilde{Z}_t\) , and \(\widetilde{I}_t = \log \mathrm {Im}(\widetilde{Z}_t)\) . We will complete the proof by determining the SDE solved by \(\widetilde{Z}_t\)  after performing a certain time change (which will follow very closely from the calculations given in the proof of [2]}). We have that
\(d \widetilde{Z}_t = -\frac{2}{\widetilde{Z}_t} dt - d\widetilde{W}_t.\) 
Then we have that
\( d \mathrm {Im}(\widetilde{Z}_t) = - \mathrm {Im}\left(\frac{2}{\widetilde{Z}_t}\right)dt.\) 
This implies that
\( d \widetilde{I}_t = d \log \mathrm {Im}( \widetilde{Z}_t ) = -\frac{1}{\mathrm {Im}(\widetilde{Z}_t)} \mathrm {Im}\left(\frac{2}{\widetilde{Z}_t} \right) = \frac{2}{|\widetilde{Z}_t|^2} dt.\) 
We also have that
\(d \log \widetilde{Z}_t &= -\frac{\sqrt{\kappa }}{\widetilde{Z}_t} d\widetilde{B}_t - \frac{2+\kappa /2}{\widetilde{Z}_t^2} dt \quad \text{and}\\d \widetilde{\Theta }_t &= -\mathrm {Im}\left( \frac{\sqrt{\kappa }}{\widetilde{Z}_t}\right) d\widetilde{B}_t - \mathrm {Im}\left( \frac{2+\kappa /2}{\widetilde{Z}_t^2} \right) dt.\) 

We now reparameterize time by letting \(ds = |\widetilde{Z}_t|^{-2} dt\) . Then we have that \(d \widetilde{I}_{t(s)} = 2 ds\)  and there exists a Brownian motion \(\widehat{B}\)  so that \(d\widehat{B}_s = d\widetilde{B}_t/|\widetilde{Z}_t|\) . Altogether, we have that
\( d \widetilde{\Theta }_{t(s)} = \sqrt{\kappa } \sin (\widetilde{\Theta }_{t(s)}) d \widehat{B}_s + \left(2+ \frac{\kappa }{2}\right) \sin (2\widetilde{\Theta }_{t(s)}) ds.\) 
In particular, it is not difficult to see for \(s > 0\)  that the law of \(\widetilde{\Theta }_{t(s)}\)  has a density with respect to Lebesgue measure. Thus if \(S \sim \exp (1)\)  independently of everything else then the law of \(\widetilde{Z}_{t(S)}\)  has a density with respect to Lebesgue measure so that \(\mathbf {P}[ \widetilde{Z}_{t(S)} \in A] = 0\) . Note that the conditional law of
\( t^{-1}(U) = \int _0^U \frac{1}{|\widetilde{Z}_u|^2} du\) 
given \(\widetilde{W}\)  is absolutely continuous with respect to Lebesgue measure on \(\mathbf {R}_+\)  as \(s \mapsto t^{-1}(s) = \int _0^s |\widetilde{Z}_u|^{-2} du\)  is a.s. strictly increasing and is a.s. \(C^1\) . Therefore the conditional law of \(t^{-1}(U)\)  given \(\widetilde{W}\)  is absolutely continuous with respect to the law of \(S\)  (which is the same as the law of \(S\)  given \(\widetilde{W}\)  by independence). Therefore the law of the pair \((\widetilde{W},t^{-1}(U))\)  is absolutely continuous with respect to the law of the pair \((\widetilde{W},S)\) . Since \(\widetilde{Z}_U\)  is obtained from \((\widetilde{W},t^{-1}(U))\)  by applying the same measurable function which constructs \(\widetilde{Z}_{t(S)}\)  from \((\widetilde{W},S)\) , this implies that the law of \(\widetilde{Z}_U\)  is absolutely continuous with respect to the law of \(\widetilde{Z}_{t(S)}\) . Therefore we also have that \(\mathbf {P}[\widetilde{Z}_U \in A] = 0\) .
